Question

If I start with 25.Og of Na and 15.0g of H2OAdd favoritea. how much H, gas will be produced?!bLR:ERC: Excess Left over:

Answer 1

Firstly, we need to balance the reaction.

The hydrogen appear in two products, so let's start with it. We need to have the same number of H in both sides, but we have 2 on the left side and 3 on the right side.

If we add a coefficient of 2 in H₂O we will have 4 H on the left side, which is now more than the 3 we have on the right side.

However, if then we add a coefficient of 2 in NaOH too, we will have 4 H in both sides.

The O will also be balanced because we will have 2 in each side.

The Na, on the other hand, we will have 1 on the left side and 2 on the irght sicde because of the coefficient 2 in NaOH, so to finish the balancing we need to add a coefficient of 2 in Na.

So, the balanced reaction is:

`2Na+2H_2O\rightarrow2NaOH+1H_2`

(a) Firslty, we need transform the masses into number of moles. For this, we will need the molar masses of Na and H₂O:

`\begin{gathered} M_(Na)=22.9898g/mol \\ M_(H₂O)=2\cdot M_H+1\cdot M_O=2\cdot1.0079g/mol+1\cdot15.9994g/mol=18.0152g/mol \end{gathered}`

So, in number of moles, we have:

`\begin{gathered} M_(Na)=(m_(Na))/(n_(Na)) \\ n_(Na)=(m_(Na))/(M_(Na))=(25.0g)/(22.9898g/mol)=1.0874...mol \end{gathered}`
`\begin{gathered} M_(H₂O)=(m_(H₂O))/(n_(H₂O)) \\ n_(H₂O)=(m_(H₂O))/(M_(H₂O))=(15.0g)/(18.0152g/mol)=0.8326...mol \end{gathered}`

Since both Na and H₂O have the same coefficients, for each Na that reacts, 1 H₂O will react, which means that we can see which one is in excess and which one is the limiting one by simple comparison.

Since we have more number of moles of Na than H₂O, we have Na in excess and H₂O is the limiting one.

So, assuming all the limiting reactant is consumed, we can calculate how much H₂ will be produced using the relation set by their coefficients:

H₂ --- H₂O

1 --- 2

`\begin{gathered} (n_(H₂))/(1)=(n_(H₂O))/(2) \\ n_(H₂)=(n_(H₂O))/(2)=(0.8326...mol)/(2)=0.4163...mol \end{gathered}`

Since the options are in grams, we need to convert this to mass, which we do as the opposite as we done with the initial masses. So we start by the molar mass of H₂:

`M_(H₂)=2\cdot M_H=2\cdot1.0079g/mol=2.0158g/mol`

Now, we covert to mass:

`\begin{gathered} M_(H₂)=(m_(H₂))/(n_(H₂)) \\ m_(H₂)=M_(H₂)n_(H₂)=2.0158g/mol\cdot0.4163...mol=0.8392g\approx0.839g \end{gathered}`

So, the mass produced of H₂ is approximately 0.839 g.

(b) Assuming LR stands for limiting reactant and ER stands for excess reactant, we already saw which is which in item (a):

LR: H₂O

ER: Na

(c) Since both Na and H₂O have the same coefficient, the number of moles of H₂O that reacted is the same as the number of moles of Na that reacted, so:

`n_(Na)=n_(H₂O)=0.8326...mol`

Since we had a total of 1.0874... mol of Na, we can calculate the number of moles left by substracting the amount that reacted:

`n_(Na,excess)=1.0874...mol-0.8326...mol=0.2548...mol`

And now we use the molar amss of Na to convert it to mass:

`\begin{gathered} M_(Na)=(m_(Na))/(n_(Na)) \\ m_(Na)=M_(Na)n_(Na)=22.9898g/mol\cdot0.2548...mol=5.8579...g\approx5.86g \end{gathered}`

We can see that we don't have the exact option, but we have the option 5.87g. The difference comes from different rounding used.

So, the excess left is approximately 5.87 g.